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1 year ago

6

Can someone show me hwo to do this so that i know how to do this please

2 answers:

spayn [35]1 year ago

5 0

Y=y coordinate (2in this case)
x=x coordinate (-2 in this case)
m=slope (-2 in this case)
b= y int. (we dont know the number because it doesn’t label it so we just put the letter and you can solve for b if the question tells you so)

so the answer is
y=mx+b (slope int. form)
2=-2(-2)+b

spin [16.1K]1 year ago

3 0

Answer:

y = -2x - 2

Step-by-step explanation:

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Shawna and her best friend Keisha go shopping. The function p(t) = 3x +2x-4x2+ 21 represents how much money each girl spent base

NNADVOKAT [17]

Answer:

$\$30$

Step-by-step explanation:

we have

$p(t)=3x+2x-4x^{2}+21$

<em>Find the amount of money that each girl spent</em>

For t= 2 hours

$p(2)=3(2)+2(2)-4(2)^{2}+21$

$p(2)=10-16+21$

$p(2)=\$15$

<em>Find the amount of money that they spend together</em>

Multiply by 2 the amount of money that each girl spent

$(2)\$15=\$30$

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3 years ago

6 radians is the same as what degree

olya-2409 [2.1K]

$1~radian=\frac{180}{\pi}degrees\\\\1~degree=\frac{\pi}{180}radians$

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3 years ago

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A dilation produces a larger figure. Which is a possible scale factor? 0 Three-fourths 1 5.

laiz [17]

Answer:

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Step-by-step explanation:

big brain and i took test

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2 years ago

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The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes

saul85 [17]

Answer:

a)0.08 , b)0.4 , C) i)0.84 , ii)0.56

Step-by-step explanation:

Given data

P(A) = professor arrives on time

P(A) = 0.8

P(B) = Student aarive on time

P(B) = 0.6

According to the question A & B are Independent

P(A∩B) = P(A) . P(B)

Therefore

${A}'$ & ${B}'$ is also independent

${A}'$ = 1-0.8 = 0.2

${B}'$ = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B') (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

$P(\frac{B'}{A})$ = $\frac{P(B'\cap A)}{P(A)}$ = $\frac{0.4\times 0.8}{0.8}$ = 0.4

Part c)

Assume the events are not independent

Given Data

P $(\frac{{A}'}{{B}'})$ = 0.4

= $\frac{P({A}'\cap {B}')}{P({B}')}$ = 0.4

$P$ $({A}'\cap {B}')$ = 0.4 x P $({B}')$

= 0.4 x 0.4 = 0.16

$P({A}'\cap {B}')$ = 0.16

i)

The probability that at least one of them is on time

$P(A\cup B)$ = 1- $P({A}'\cap {B}')$

= 1 - 0.16 = 0.84

ii)The probability that they are both on time

P $(A\cap B)$ = 1 - $P({A}'\cup {B}')$ = 1 - $[P({A}')+P({B}') - P({A}'\cap {B}')]$

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3 years ago

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beks73 [17]

I think the answer is B

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2 years ago

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